Solve the differential equation 1 + x 2 d y d x + y = e tan - 1 x , given that y = 1 when x = 0 .

We have, 1 + x 2 d y d x + y = e tan - 1 x ⇒ d y d x + y 1 + x 2 = e tan - 1 x 1 + x 2 Comparing it with standard equation, d y d x + P y = Q , where P = 1 1 + x 2 and Q = e tan - 1 x 1 + x 2 Now, integrating factor I.F = e ∫ P d x = e ∫ 1 1 + x 2 d x = e tan - 1 x So, the solution is given by: y I . F = ∫ Q x I . F d x + c ⇒ y e tan - 1 x = ∫ e tan - 1 x 1 + x 2 · e tan - 1 x d x + c . . . 1 Let I = ∫ e tan - 1 x 1 + x 2 · e tan - 1 x d x Putting e tan - 1 x = t ⇒ 1 1 + x 2 · e tan - 1 x d x = d t Thus, I = ∫ t d t = t 2 2 + c ⇒ I = e tan - 1 x 2 2 + c Considering 1 , we get: y e tan - 1 x = e tan - 1 x 2 2 + c It is given that y = 1 , x = 0 ⇒ 1 · e tan - 1 0 = e tan - 1 0 2 2 + c ⇒ 1 · e tan - 1 tan 0 = e tan - 1 tan 0 2 2 + c ⇒ 1 · e 0 = e 0 2 2 + c ∵ tan - 1 tan A = A ⇒ 1 = 1 4 + c ⇒ c = 3 4 Thus, the solution is y e tan - 1 x = e tan - 1 x 2 2 + 3 4 .

Find the integrating factor of the linear equation d y d x + P y = Q and hence obtain the general solution of the equation.

A differential equation is linear if the dependent variable y and its derivative appear only in first degree. The general form of a linear differential equation of first order is d y d x + P y = Q . . . 1 where P and Q are functions of x (or constants) This type of differential equations are solved when they are multiplied by a factor, which is called the integrating factor, because by multiplication of this factor the LHS of differential equation 1 becomes exact differential of some function. Multiplying both sides of 1 by e ∫ P d x , we get e ∫ P d x d y d x + P y = Q e ∫ P d x ⇒ d d x y e ∫ P d x = Q e ∫ P d x On integrating both sides w.r.t x , we get y e ∫ P d x = ∫ Q e ∫ P d x + C . . . 2 This is the required solution, where C is the constant of integration. Here, e ∫ P d x is called the integrating factor. The solution 2 in short may also be written as y · I . F . = ∫ Q · I . F . d x + C

Find the integrating factor of the linear equation d y d x + P y = Q and hence obtain the general solution of the equation.

A differential equation is linear if the dependent variable y and its derivative appear only in first degree. The general form of a linear differential equation of first order is d y d x + P y = Q . . . 1 where P and Q are functions of x (or constants) This type of differential equations are solved when they are multiplied by a factor, which is called the integrating factor, because by multiplication of this factor the LHS of differential equation 1 becomes exact differential of some function. Multiplying both sides of 1 by e ∫ P d x , we get e ∫ P d x d y d x + P y = Q e ∫ P d x ⇒ d d x y e ∫ P d x = Q e ∫ P d x On integrating both sides w.r.t x , we get y e ∫ P d x = ∫ Q e ∫ P d x + C . . . 2 This is the required solution, where C is the constant of integration. Here, e ∫ P d x is called the integrating factor. The solution 2 in short may also be written as y · I . F . = ∫ Q · I . F . d x + C

Solve the differential equation 1 + x 2 d y d x + y = e tan - 1 x , given that y = 1 when x = 0 .

We have, 1 + x 2 d y d x + y = e tan - 1 x ⇒ d y d x + y 1 + x 2 = e tan - 1 x 1 + x 2 Comparing it with standard equation, d y d x + P y = Q , where P = 1 1 + x 2 and Q = e tan - 1 x 1 + x 2 Now, integrating factor I.F = e ∫ P d x = e ∫ 1 1 + x 2 d x = e tan - 1 x So, the solution is given by: y I . F = ∫ Q x I . F d x + c ⇒ y e tan - 1 x = ∫ e tan - 1 x 1 + x 2 · e tan - 1 x d x + c . . . 1 Let I = ∫ e tan - 1 x 1 + x 2 · e tan - 1 x d x Putting e tan - 1 x = t ⇒ 1 1 + x 2 · e tan - 1 x d x = d t Thus, I = ∫ t d t = t 2 2 + c ⇒ I = e tan - 1 x 2 2 + c Considering 1 , we get: y e tan - 1 x = e tan - 1 x 2 2 + c It is given that y = 1 , x = 0 ⇒ 1 · e tan - 1 0 = e tan - 1 0 2 2 + c ⇒ 1 · e tan - 1 tan 0 = e tan - 1 tan 0 2 2 + c ⇒ 1 · e 0 = e 0 2 2 + c ∵ tan - 1 tan A = A ⇒ 1 = 1 4 + c ⇒ c = 3 4 Thus, the solution is y e tan - 1 x = e tan - 1 x 2 2 + 3 4 .

Show that the general solution of the differential equation d y d x + y 2 + y + 1 x 2 + x + 1 = 0 is ( x + y + 1 ) = A ( 1 − x − y − 2 x y ) , where A is an arbitrary constant.

Given differential equation is d y d x + y 2 + y + 1 x 2 + x + 1 = 0 ⇒ d y d x = - y 2 + y + 1 x 2 + x + 1 ⇒ d y y 2 + y + 1 = - d x x 2 + x + 1 ⇒ d y y 2 + y + 1 + d x x 2 + x + 1 = 0 Integrating both sides, we get: ∫ d y y 2 + y + 1 + ∫ d x x 2 + x + 1 = C ⇒ ∫ d y y + 1 2 2 + 3 2 2 + ∫ d x x + 1 2 2 + 3 2 2 = C ⇒ 2 3 t a n - 1 y + 1 2 3 2 + 2 3 t a n - 1 x + 1 2 3 2 = C ⇒ t a n - 1 2 y + 1 3 + t a n - 1 2 x + 1 3 = 3 C 2 ⇒ t a n - 1 2 y + 1 3 + 2 x + 1 3 1 - 2 y + 1 3 · 2 x + 1 3 = 3 C 2 ⇒ t a n - 1 2 x + 2 y + 2 3 1 - 4 x y + 2 x + 2 y + 1 3 = 3 C 2 ⇒ t a n - 1 2 3 x + y + 1 3 - 4 x y - 2 x - 2 y - 1 = 3 C 2 ⇒ t a n - 1 3 x + y + 1 2 1 - x - y - 2 x y = 3 C 2 ⇒ 3 x + y + 1 2 1 - x - y - 2 x y = tan ⁡ 3 C 2 = B , where B = tan ⁡ 3 C 2 ⇒ x + y + 1 = 2 B 3 1 - x - y - 2 x y ⇒ x + y + 1 = A 1 - x - y - 2 x y , where A = 2 B 3 Hence proved